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            <h1 id="seo-header">『算法-ACM竞赛-CodeForces』1255Round#601(Div.2)C.LeagueofLeesins（大模拟）</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-CodeForces』1255Round-601-Div-2-C-LeagueofLeesins（大模拟）"><a href="#『算法-ACM-竞赛-CodeForces』1255Round-601-Div-2-C-LeagueofLeesins（大模拟）" class="headerlink" title="『算法-ACM 竞赛-CodeForces』1255Round#601(Div.2)C.LeagueofLeesins（大模拟）"></a>『算法-ACM 竞赛-CodeForces』1255Round#601(Div.2)C.LeagueofLeesins（大模拟）</h1><h1 id="Codeforce-1255-Round-601-Div-2-C-League-of-Leesins-（大模拟）"><a href="#Codeforce-1255-Round-601-Div-2-C-League-of-Leesins-（大模拟）" class="headerlink" title="Codeforce 1255 Round #601 (Div. 2) C. League of Leesins （大模拟）"></a>Codeforce 1255 Round #601 (Div. 2) C. League of Leesins （大模拟）</h1><p>Bob is an avid fan of the video game “League of Leesins”, and today he celebrates as the League of Leesins World Championship comes to an end!</p>
<p>The tournament consisted of nn (n≥5n≥5) teams around the world. Before the tournament starts, Bob has made a prediction of the rankings of each team, from 11-st to nn-th. After the final, he compared the prediction with the actual result and found out that the ii-th team according to his prediction ended up at the pipi-th position (1≤pi≤n1≤pi≤n, all pipi are unique). In other words, pp is a permutation of 1,2,…,n1,2,…,n.</p>
<p>As Bob’s favorite League player is the famous “3ga”, he decided to write down every 33 consecutive elements of the permutation pp. Formally, Bob created an array qq of n−2n−2 triples, where qi&#x3D;(pi,pi+1,pi+2)qi&#x3D;(pi,pi+1,pi+2) for each 1≤i≤n−21≤i≤n−2. Bob was very proud of his array, so he showed it to his friend Alice.</p>
<p>After learning of Bob’s array, Alice declared that she could retrieve the permutation pp even if Bob rearranges the elements of qq and the elements within each triple. Of course, Bob did not believe in such magic, so he did just the same as above to see Alice’s respond.</p>
<p>For example, if n&#x3D;5n&#x3D;5 and p&#x3D;[1,4,2,3,5]p&#x3D;[1,4,2,3,5], then the original array qq will be [(1,4,2),(4,2,3),(2,3,5)][(1,4,2),(4,2,3),(2,3,5)]. Bob can then rearrange the numbers within each triple and the positions of the triples to get [(4,3,2),(2,3,5),(4,1,2)][(4,3,2),(2,3,5),(4,1,2)]. Note that [(1,4,2),(4,2,2),(3,3,5)][(1,4,2),(4,2,2),(3,3,5)] is not a valid rearrangement of qq, as Bob is not allowed to swap numbers belong to different triples.</p>
<p>As Alice’s friend, you know for sure that Alice was just trying to show off, so you decided to save her some face by giving her any permutation pp that is consistent with the array qq she was given.</p>
<p>Input</p>
<p>The first line contains a single integer nn (5≤n≤1055≤n≤105) — the size of permutation pp.</p>
<p>The ii-th of the next n−2n−2 lines contains 33 integers qi,1qi,1, qi,2qi,2, qi,3qi,3 (1≤qi,j≤n1≤qi,j≤n) — the elements of the ii-th triple of the rearranged (shuffled) array qiqi, in random order. Remember, that the numbers within each triple can be rearranged and also the positions of the triples can be rearranged.</p>
<p>It is guaranteed that there is at least one permutation pp that is consistent with the input.</p>
<p>Output</p>
<p>Print nn distinct integers p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n) such that pp is consistent with array qq.</p>
<p>If there are multiple answers, print any.</p>
<p>Example</p>
<p>Input</p>
<pre><code class="hljs">5
4 3 2
2 3 5
4 1 2
</code></pre>
<p>Output</p>
<pre><code class="hljs">1 4 2 3 5
</code></pre>
<p>这个题，出现次数最少的在两边，确定了第一个为出现次数 1 2 3 的那一组，然后根据那一组暴力即可，用 STL 优化了一下，不然会超时。</p>
<pre><code class="hljs">#include&lt;bits/stdc++.h&gt;
using namespace std;
int a[100000],b[100000],c[100000],vis[1000000],n;
vector&lt;int&gt; v[1000000];
int main()
&#123;
    cin&gt;&gt;n;
    for(int i=0;i&lt;n-2;++i)
    &#123;
        cin&gt;&gt;a[i]&gt;&gt;b[i]&gt;&gt;c[i];
        v[a[i]-1].push_back(i);
        v[b[i]-1].push_back(i);
        v[c[i]-1].push_back(i);
        vis[a[i]-1]=1;
        vis[b[i]-1]=1;
        vis[c[i]-1]=1;
    &#125;
    int x=0;
    for(;v[x].size()!=1;++x);
        vis[x]=0;
    cout&lt;&lt;x+1&lt;&lt;&quot; &quot;;
    int y,z;
    if(v[a[v[x][0]]-1].size()==2)
    &#123;
        y=a[v[x][0]]-1;
        vis[y]=0;
    &#125;
    else if(v[a[v[x][0]]-1].size()==3)
    &#123;
        z=a[v[x][0]]-1;
        vis[z]=0;
    &#125;
    if(v[b[v[x][0]]-1].size()==2)
    &#123;
        y=b[v[x][0]]-1;
        vis[y]=0;
    &#125;
    else if(v[b[v[x][0]]-1].size()==3)
    &#123;
        z=b[v[x][0]]-1;
        vis[z]=0;	&#125;
    if(v[c[v[x][0]]-1].size()==2)
    &#123;
        y=c[v[x][0]]-1;
        vis[y]=0;
    &#125;
    else if(v[c[v[x][0]]-1].size()==3)
    &#123;
        z=c[v[x][0]]-1;
        vis[z]=0;
    &#125;
    cout&lt;&lt;y+1&lt;&lt;&quot; &quot;&lt;&lt;z+1&lt;&lt;&quot; &quot;;
    int h=z;
    int cnt=3;
    while(cnt++&lt;n)
    &#123;
        for(int i=0;i&lt;v[h].size();++i)
        &#123;

            if(vis[a[v[h][i]]-1]+vis[b[v[h][i]]-1]+vis[c[v[h][i]]-1]==1)
            &#123;
                if(vis[a[v[h][i]]-1]==1)&#123;
                    cout&lt;&lt;a[v[h][i]]&lt;&lt;&quot; &quot;;
                    vis[a[v[h][i]]-1]=0;
                    h=a[v[h][i]]-1;
                &#125;
                else if(vis[b[v[h][i]]-1]==1)&#123;
                    cout&lt;&lt;b[v[h][i]]&lt;&lt;&quot; &quot;;
                    vis[b[v[h][i]]-1]=0;
                    h=b[v[h][i]]-1;
                &#125;
                else if(vis[c[v[h][i]]-1]==1)&#123;
                    cout&lt;&lt;c[v[h][i]]&lt;&lt;&quot; &quot;;
                    vis[c[v[h][i]]-1]=0;
                    h=c[v[h][i]]-1;
                &#125;
                break;
            &#125;
        &#125;
    &#125;
&#125;
</code></pre>

                
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